Given two arrays nums1 and nums2.
Return the maximum dot product between non-empty subsequences of nums1 and nums2 with the same length.
A subsequence of a array is a new array which is formed from the original array by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, [2,3,5] is a subsequence of [1,2,3,4,5] while [1,5,3] is not).
Example 1:
Input: nums1 = [2,1,-2,5], nums2 = [3,0,-6] Output: 18 Explanation: Take subsequence [2,-2] from nums1 and subsequence [3,-6] from nums2. Their dot product is (2*3 + (-2)*(-6)) = 18.
Example 2:
Input: nums1 = [3,-2], nums2 = [2,-6,7] Output: 21 Explanation: Take subsequence [3] from nums1 and subsequence [7] from nums2. Their dot product is (3*7) = 21.
Example 3:
Input: nums1 = [-1,-1], nums2 = [1,1] Output: -1 Explanation: Take subsequence [-1] from nums1 and subsequence [1] from nums2. Their dot product is -1.
Constraints:
1 <= nums1.length, nums2.length <= 500-1000 <= nums1[i], nums2[i] <= 1000
Solution: DP
dp[i][j] := max product of nums1[0~i], nums2[0~j].
dp[i][j] = max(dp[i-1][j], dp[i][j -1], max(0, dp[i-1][j-1]) + nums1[i]*nums2[j])
Time complexity: O(n1*n2)
Space complexity: O(n1*n2)
C++
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// Author: Huahua class Solution { public: int maxDotProduct(vector<int>& nums1, vector<int>& nums2) { const int n1 = nums1.size(); const int n2 = nums2.size(); // dp[i][j] = max dot product of two subsequences // of nums1[0:i] and nums2[0:j] vector<vector<int>> dp(n1 + 1, vector<int>(n2 + 1, INT_MIN / 2)); for (int i = 1; i <= n1; ++i) for (int j = 1; j <= n2; ++j) dp[i][j] = max({ dp[i - 1][j], dp[i][j - 1], max(0, dp[i - 1][j - 1]) + nums1[i - 1] * nums2[j - 1] }); return dp[n1][n2]; } }; |
C++
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// Author: Huahua class Solution { public: int maxDotProduct(vector<int>& nums1, vector<int>& nums2) { const int n1 = nums1.size(); const int n2 = nums2.size(); // dp[i][j] = max dot product of two subsequences // of nums1[0~i] and nums2[0~j] vector<vector<int>> dp(n1, vector<int>(n2)); for (int i = 0; i < n1; ++i) for (int j = 0; j < n2; ++j) { dp[i][j] = nums1[i] * nums2[j]; if (i > 0 && j > 0) dp[i][j] += max(0, dp[i - 1][j - 1]); if (i > 0) dp[i][j] = max(dp[i][j], dp[i - 1][j]); if (j > 0) dp[i][j] = max(dp[i][j], dp[i][j - 1]); } return dp.back().back(); } }; |
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